Exercise 1: The power rule, after rewriting roots and fractions as powers

The power rule $\frac{d}{dx}x^n = nx^{n-1}$ holds for EVERY real exponent $n$ : positive, negative, fractional, even irrational. But it applies to one form only, a power of $x$ with a constant exponent. A root, a fraction with $x$ in its denominator, or a quotient by a power of $x$ must first be REWRITTEN as a sum of terms $cx^n$ ; only then is the rule quoted, term by term, with the sum and constant multiple rules.

Two more facts complete the toolkit: the derivative of a constant is $0$ , whatever the constant looks like ( $\pi^3$ , $e^2$ , $\sqrt 7$ ), and $\frac{d}{dx}e^x = e^x$ .

a) Differentiate $f(x) = 4x^5 - 3x^2 + 7x - \pi^3$ .
b) Differentiate $g(x) = 6\sqrt[3]{x} - \frac{8}{\sqrt x} + \frac{5}{x^2}$ , and give the answer without negative or fractional exponents.
c) Let $h(x) = \frac{x^2 - 3x + 5}{\sqrt x}$ for $x > 0$ . Find $h'(x)$ as a single fraction, then $h'(4)$ .
d) Differentiate $m(x) = x^e + e^x + e^e$ and $p(x) = (2\sqrt x + 1)^2$ , the second without the chain rule.
e) A student writes: $\frac{d}{dx}\left(\frac{1}{3x^2}\right) = \frac{1}{6x}$ . Name the two errors and give the correct derivative.

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Answers

a) $f'(x) = 20x^4 - 6x + 7$
b) $g'(x) = \frac{2}{\sqrt[3]{x^2}} + \frac{4}{x\sqrt x} - \frac{10}{x^3}$
c) $h'(x) = \frac{3x^2 - 3x - 5}{2x\sqrt x}$ , $h'(4) = \frac{31}{16}$
d) $m'(x) = ex^{e-1} + e^x$ ; $p'(x) = 4 + \frac{2}{\sqrt x}$
e) The denominator was differentiated and the fraction kept: $\frac{d}{dx}\left(\frac{1}{3}x^{-2}\right) = -\frac{2}{3x^3}$

a) By the sum and constant multiple rules, each term is differentiated separately: $(4x^5)' = 4 \cdot 5x^4 = 20x^4$ , $(-3x^2)' = -6x$ , $(7x)' = 7$ , and $(\pi^3)' = 0$ because $\pi^3 \approx 31$ is a NUMBER, not a power of $x$ . So $f'(x) = 20x^4 - 6x + 7$ . Writing $3\pi^2$ for the derivative of $\pi^3$ applies the power rule to a constant: the variable is $x$ , and $\pi^3$ does not contain it. That single slip costs a mark on every paper where it appears.

b) Rewrite first: $\sqrt[3]{x} = x^{1/3}$ , $\frac{1}{\sqrt x} = x^{-1/2}$ and $\frac{1}{x^2} = x^{-2}$ , so $g(x) = 6x^{1/3} - 8x^{-1/2} + 5x^{-2}$ . The power rule then gives $g'(x) = 6 \cdot \frac{1}{3}x^{-2/3} - 8 \cdot \left(-\frac{1}{2}\right)x^{-3/2} + 5 \cdot (-2)x^{-3} = 2x^{-2/3} + 4x^{-3/2} - 10x^{-3}$ . Back to radicals: $g'(x) = \frac{2}{\sqrt[3]{x^2}} + \frac{4}{x\sqrt x} - \frac{10}{x^3}$ , since $x^{3/2} = x\sqrt x$ . The trap is the sign of the middle term: the exponent $-\frac{1}{2}$ multiplies $-8$ , and two minus signs make $+4$ . Subtracting $1$ from a negative exponent also moves it AWAY from zero: $-\frac{1}{2} - 1 = -\frac{3}{2}$ , not $-\frac{1}{2} + 1$ .

c) The denominator is a single power of $x$ , so divide term by term instead of using the quotient rule: $h(x) = \frac{x^2}{x^{1/2}} - \frac{3x}{x^{1/2}} + \frac{5}{x^{1/2}} = x^{3/2} - 3x^{1/2} + 5x^{-1/2}$ . Then $h'(x) = \frac{3}{2}x^{1/2} - \frac{3}{2}x^{-1/2} - \frac{5}{2}x^{-3/2}$ . Over the common denominator $2x^{3/2} = 2x\sqrt x$ : $\frac{3}{2}x^{1/2} = \frac{3x^2}{2x^{3/2}}$ and $\frac{3}{2}x^{-1/2} = \frac{3x}{2x^{3/2}}$ , so $h'(x) = \frac{3x^2 - 3x - 5}{2x\sqrt x}$ . At $x = 4$ : $\sqrt 4 = 2$ , so $h'(4) = \frac{48 - 12 - 5}{2 \cdot 4 \cdot 2} = \frac{31}{16}$ . The quotient rule would give the same answer after twice the algebra; rewriting is the method a marker expects when the denominator is a monomial.

d) The three terms of $m$ look alike and are three different objects. $x^e$ is a POWER of $x$ with the constant exponent $e$ : $(x^e)' = ex^{e-1}$ . $e^x$ is the exponential, constant base and variable exponent: $(e^x)' = e^x$ , and the power rule does not apply to it. $e^e$ is a constant: its derivative is $0$ . So $m'(x) = ex^{e-1} + e^x$ . For $p$ , expand the square, which removes any composition: $p(x) = 4x + 4\sqrt x + 1 = 4x + 4x^{1/2} + 1$ , so $p'(x) = 4 + 2x^{-1/2} = 4 + \frac{2}{\sqrt x}$ . Check at $x = 1$ : $p$ goes from $p(1) = 9$ with slope $6$ , and indeed $4 + 2 = 6$ .

e) First error: the student differentiated the denominator $3x^2$ into $6x$ and kept the fraction bar, as if $\left(\frac{1}{u}\right)' = \frac{1}{u'}$ , a rule that does not exist. Second error, as a consequence: the sign is lost, although $\frac{1}{3x^2}$ is decreasing for $x > 0$ , so its derivative there must be NEGATIVE. The correct route rewrites first: $\frac{1}{3x^2} = \frac{1}{3}x^{-2}$ , so the derivative is $\frac{1}{3} \cdot (-2)x^{-3} = -\frac{2}{3x^3}$ . At $x = 1$ the student's answer is $\frac{1}{6}$ and the true slope is $-\frac{2}{3}$ : one test value is enough to expose the rule.

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Exercise 2: Product and quotient rules: never factor by factor

The derivative of a product is NOT the product of the derivatives, and the derivative of a quotient is not the quotient of the derivatives. The rules are $(fg)' = f'g + fg'$ and $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ . In the quotient rule the ORDER of the numerator matters, because of the minus sign, and the $g^2$ of the denominator is left as it is: expanding it only produces a longer expression that no longer factors.

In every part, name the rule you are applying and the two functions $f$ and $g$ before you write the derivative.

a) Differentiate $F(x) = (x^2 + 1)(x^3 - 2x)$ with the product rule, then check your answer by expanding $F$ first.
b) Differentiate $G(x) = \frac{3x - 1}{x^2 + 2}$ and compute $G'(0)$ .
c) Differentiate $H(x) = x^2e^x$ , factor the result, and find the values of $x$ where $H'(x) = 0$ .
d) Differentiate $K(x) = \frac{1 + e^x}{1 - e^x}$ for $x \ne 0$ , and simplify.
e) A student writes $(x^2e^x)' = 2xe^x$ and $\left(\frac{x}{x+1}\right)' = \frac{1}{1} = 1$ . Refute both answers with a single value of $x$ each.

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Answers

a) $F'(x) = 5x^4 - 3x^2 - 2$
b) $G'(x) = \frac{-3x^2 + 2x + 6}{(x^2 + 2)^2}$ , $G'(0) = \frac{3}{2}$
c) $H'(x) = x(x + 2)e^x$ , zero at $x = 0$ and $x = -2$
d) $K'(x) = \frac{2e^x}{(1 - e^x)^2}$
e) At $x = 1$ : $3e \ne 2e$ ; at $x = 1$ : $\frac{1}{4} \ne 1$

a) Product rule with $f(x) = x^2 + 1$ , $f'(x) = 2x$ , and $g(x) = x^3 - 2x$ , $g'(x) = 3x^2 - 2$ : $F'(x) = 2x(x^3 - 2x) + (x^2 + 1)(3x^2 - 2) = 2x^4 - 4x^2 + 3x^4 - 2x^2 + 3x^2 - 2 = 5x^4 - 3x^2 - 2$ . Check by expanding first: $F(x) = x^5 - 2x^3 + x^3 - 2x = x^5 - x^3 - 2x$ , so $F'(x) = 5x^4 - 3x^2 - 2$ , the same. On a product of two polynomials both routes are legal; expanding is the natural CHECK of the product rule, and the factor-by-factor answer $2x(3x^2 - 2)$ fails it at once, being of degree $3$ instead of $4$ .

b) Quotient rule with $f(x) = 3x - 1$ , $f'(x) = 3$ , and $g(x) = x^2 + 2$ , $g'(x) = 2x$ : $G'(x) = \frac{3(x^2 + 2) - (3x - 1)(2x)}{(x^2 + 2)^2} = \frac{3x^2 + 6 - 6x^2 + 2x}{(x^2 + 2)^2} = \frac{-3x^2 + 2x + 6}{(x^2 + 2)^2}$ . Then $G'(0) = \frac{6}{4} = \frac{3}{2}$ . The parentheses around $(3x - 1)(2x)$ carry the minus sign to BOTH terms: dropping them gives $-6x^2 - 2x$ and a wrong numerator. Writing $fg' - f'g$ in the numerator flips the sign of the whole answer; the order follows the rule's reading, derivative of the TOP first.

c) Product rule with $f(x) = x^2$ and $g(x) = e^x$ , whose derivative is $e^x$ itself: $H'(x) = 2x \cdot e^x + x^2 \cdot e^x = (x^2 + 2x)e^x = x(x + 2)e^x$ . Since $e^x > 0$ for every $x$ , the product is zero exactly when $x(x + 2) = 0$ : $x = 0$ or $x = -2$ . Factoring $e^x$ out is not cosmetic: it is the step that turns an equation with an exponential into a polynomial equation, and the fact $e^x \ne 0$ must be SAID to justify dividing by it.

d) Quotient rule with $f(x) = 1 + e^x$ , $f'(x) = e^x$ , and $g(x) = 1 - e^x$ , $g'(x) = -e^x$ : $K'(x) = \frac{e^x(1 - e^x) - (1 + e^x)(-e^x)}{(1 - e^x)^2}$ . The numerator is $e^x - e^xe^x + e^x + e^xe^x = 2e^x$ , the two products $e^xe^x$ cancelling. So $K'(x) = \frac{2e^x}{(1 - e^x)^2}$ , positive wherever it is defined. The double minus in $-(1 + e^x)(-e^x)$ is where most papers lose the answer: write the product $fg'$ in full, with its own sign, before expanding.

e) At $x = 1$ , the true derivative from c) is $H'(1) = 1 \cdot 3 \cdot e = 3e$ , while $2xe^x$ gives $2e$ : the student multiplied the derivatives of the two factors and forgot the term $x^2e^x$ . For the quotient, the true derivative is $\frac{1 \cdot (x + 1) - x \cdot 1}{(x + 1)^2} = \frac{1}{(x + 1)^2}$ , which equals $\frac{1}{4}$ at $x = 1$ , not $1$ . Choosing the test value matters: at $x = 0$ both wrong answers happen to agree with the right ones ( $0$ and $1$ ), so a single lucky value proves nothing. A test value can refute a rule, never confirm it.

Exercise 3: Choosing the form: rewrite first, or use the quotient rule?

The quotient rule always works on a quotient, but it is often the longest road. Three shapes deserve a shortcut: a denominator that is a single power of $x$ (divide term by term), a CONSTANT numerator (the reciprocal rule $\left(\frac{c}{g}\right)' = -\frac{cg'}{g^2}$ , a special case of the quotient rule), and a quotient that simplifies, provided the domain is kept. The laws of exponents play the same role for $e^x$ : without the chain rule, which comes two chapters later, $e^{x+2}$ , $e^{-x}$ and $e^{2x}$ can only be differentiated after being rewritten.

The figure shows the graph of $g(x) = \frac{x^2 - 9}{x - 3}$ .

a) Differentiate $f(x) = \frac{x^3 - 4x + 1}{x^2}$ in two ways, by rewriting and by the quotient rule, and check that the answers agree.
b) Differentiate $q(x) = \frac{5}{x^2 + 4}$ and compute $q'(1)$ . Why is $\frac{(5)'}{(x^2 + 4)'} = 0$ absurd?
c) Find $g'(x)$ for the function of the figure. What happens at $x = 3$ ?
d) Without the chain rule, differentiate $e^{x+2}$ , $e^{-x}$ and $e^{2x}$ . A student claims $(e^{-x})' = e^{-x}$ , since $e^x$ is its own derivative: refute it by looking at the graph of $e^{-x}$ .
e) Differentiate $w(x) = \frac{x^2 + 1}{x^2 - 1}$ by first writing it as $1 + \frac{c}{x^2 - 1}$ , and check with the quotient rule.

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Answers

a) $f'(x) = 1 + \frac{4}{x^2} - \frac{2}{x^3} = \frac{x^3 + 4x - 2}{x^3}$
b) $q'(x) = -\frac{10x}{(x^2 + 4)^2}$ , $q'(1) = -\frac{2}{5}$ ; $q$ is not constant, so its derivative is not $0$
c) $g'(x) = 1$ for $x \ne 3$ ; $g'(3)$ does not exist, $g$ is not defined at $3$
d) $(e^{x+2})' = e^{x+2}$ , $(e^{-x})' = -e^{-x}$ , $(e^{2x})' = 2e^{2x}$ ; $e^{-x}$ decreases, so its slope is negative
e) $w(x) = 1 + \frac{2}{x^2 - 1}$ , $w'(x) = -\frac{4x}{(x^2 - 1)^2}$

a) Rewriting: $f(x) = \frac{x^3}{x^2} - \frac{4x}{x^2} + \frac{1}{x^2} = x - 4x^{-1} + x^{-2}$ , so $f'(x) = 1 + 4x^{-2} - 2x^{-3} = 1 + \frac{4}{x^2} - \frac{2}{x^3}$ . Quotient rule, with $(x^3 - 4x + 1)' = 3x^2 - 4$ and $(x^2)' = 2x$ : $f'(x) = \frac{(3x^2 - 4)x^2 - (x^3 - 4x + 1)(2x)}{x^4} = \frac{3x^4 - 4x^2 - 2x^4 + 8x^2 - 2x}{x^4} = \frac{x^4 + 4x^2 - 2x}{x^4} = \frac{x^3 + 4x - 2}{x^3}$ after dividing by $x$ . Over the denominator $x^3$ , the first answer is also $\frac{x^3 + 4x - 2}{x^3}$ : they agree. The rewriting took one line, the quotient rule four, with a denominator $x^4$ to simplify at the end.

b) Reciprocal rule with $c = 5$ and $g(x) = x^2 + 4$ : $q'(x) = -\frac{5 \cdot 2x}{(x^2 + 4)^2} = -\frac{10x}{(x^2 + 4)^2}$ , and $q'(1) = -\frac{10}{25} = -\frac{2}{5}$ . The quotient rule gives the same, since the $f'g$ term vanishes: $\frac{0 \cdot (x^2 + 4) - 5 \cdot 2x}{(x^2 + 4)^2}$ . The answer $\frac{0}{2x} = 0$ would mean that $q$ has a horizontal tangent everywhere, so that $q$ is constant; yet $q(0) = \frac{5}{4}$ and $q(1) = 1$ . A constant NUMERATOR does not make the quotient constant.

c) For $x \ne 3$ , $x^2 - 9 = (x - 3)(x + 3)$ and the factor $x - 3$ cancels: $g(x) = x + 3$ , so $g'(x) = 1$ . This is exactly the figure, a line of slope $1$ . But $g$ is not defined at $x = 3$ , which the open circle at $(3, 6)$ shows, so $g'(3)$ does not exist: a derivative at a point needs the function AT that point. Check with the quotient rule: $\frac{2x(x - 3) - (x^2 - 9)}{(x - 3)^2} = \frac{x^2 - 6x + 9}{(x - 3)^2} = 1$ for $x \ne 3$ . Simplifying is allowed and fast, as long as the answer keeps the condition $x \ne 3$ that the simplification hides.

d) By the laws of exponents, $e^{x+2} = e^2e^x$ , a constant times $e^x$ : its derivative is $e^2e^x = e^{x+2}$ . Next, $e^{-x} = \frac{1}{e^x}$ , so by the reciprocal rule $(e^{-x})' = -\frac{e^x}{(e^x)^2} = -\frac{1}{e^x} = -e^{-x}$ . Finally $e^{2x} = e^xe^x$ , and the product rule gives $e^xe^x + e^xe^x = 2e^{2x}$ . The student's claim is refuted by one look at the graph of $e^{-x}$ : it DECREASES on the whole real line, so its tangent lines all have negative slopes, and a derivative equal to $e^{-x} > 0$ is impossible. The rule $(e^x)' = e^x$ is about the exponent $x$ exactly; any other exponent must be rewritten first here, or handled by the chain rule later.

e) Divide first: $x^2 + 1 = (x^2 - 1) + 2$ , so $w(x) = 1 + \frac{2}{x^2 - 1}$ , for $x \ne \pm 1$ . The constant $1$ has derivative $0$ and the reciprocal rule gives $w'(x) = -\frac{2 \cdot 2x}{(x^2 - 1)^2} = -\frac{4x}{(x^2 - 1)^2}$ . Check with the quotient rule: $\frac{2x(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2}$ . Same answer. When numerator and denominator have the same degree, one division turns the quotient into a constant plus a simpler fraction.

Exercise 4: Tangent and normal lines, and where the tangent is horizontal

The tangent line to $y = f(x)$ at $x = a$ is $y = f(a) + f'(a)(x - a)$ : a line, so its slope $f'(a)$ is a NUMBER, obtained by differentiating first and substituting $a$ afterwards. The normal line at the same point is perpendicular to the tangent: when $f'(a) \ne 0$ its slope is $-\frac{1}{f'(a)}$ , and when $f'(a) = 0$ the tangent is horizontal and the normal is the vertical line $x = a$ .

The figure shows the graph of $f(x) = x^3 - 3x^2 - 9x + 5$ on $[-3, 5]$ .

a) Find $f'(x)$ , then the equation of the tangent line at the point of abscissa $1$ , in the form $y = mx + b$ .
b) Find the equation of the normal line at the same point.
c) Find the points of the curve where the tangent is horizontal, and give the normal line at each of them.
d) Find the points where the tangent is parallel to the line $y = 15x + 2$ .
e) Write $f'(x)$ in the form $3(x - 1)^2 - k$ . Deduce that no tangent to the curve has slope $-13$ , and find the smallest slope of a tangent.

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Answers

a) $f'(x) = 3x^2 - 6x - 9$ ; tangent $y = -12x + 6$
b) Normal $y = \frac{1}{12}x - \frac{73}{12}$
c) $(-1, 10)$ and $(3, -22)$ ; normals $x = -1$ and $x = 3$
d) $(-2, 3)$ and $(4, -15)$
e) $f'(x) = 3(x - 1)^2 - 12 \ge -12$ : no slope $-13$ ; smallest slope $-12$ , at $x = 1$

a) By the power, sum and constant multiple rules, $f'(x) = 3x^2 - 6x - 9$ . At $a = 1$ : $f(1) = 1 - 3 - 9 + 5 = -6$ and $f'(1) = 3 - 6 - 9 = -12$ . The tangent is $y = -6 - 12(x - 1) = -12x + 6$ . Check: at $x = 1$ the line gives $-12 + 6 = -6 = f(1)$ , so it does pass through the point. On the figure the curve is indeed going down steeply at $x = 1$ . The form $y = -6 + (3x^2 - 6x - 9)(x - 1)$ , with $f'(x)$ left unevaluated, is a CUBIC, not a line, and earns no mark: substitute $a$ into $f'$ before writing the equation.

b) The tangent has slope $-12 \ne 0$ , so the normal has slope $-\frac{1}{-12} = \frac{1}{12}$ and passes through $(1, -6)$ : $y = -6 + \frac{1}{12}(x - 1) = \frac{1}{12}x - \frac{73}{12}$ . The check is the product of the slopes: $(-12) \cdot \frac{1}{12} = -1$ , the condition for two non-vertical lines to be perpendicular. Answering with slope $12$ (the opposite, forgetting the reciprocal) or $-\frac{1}{12}$ (the reciprocal, forgetting the sign) are the two classic errors, and each costs the question.

c) A horizontal tangent means slope $0$ : $3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) = 0$ , so $x = -1$ or $x = 3$ . Then $f(-1) = -1 - 3 + 9 + 5 = 10$ and $f(3) = 27 - 27 - 27 + 5 = -22$ : the points are $(-1, 10)$ and $(3, -22)$ , the top of the hump and the bottom of the valley on the figure. At each of them the tangent is horizontal, $y = 10$ and $y = -22$ , and the normal is VERTICAL: $x = -1$ and $x = 3$ . The formula $-\frac{1}{f'(a)}$ breaks down here, and writing no normal exists is wrong: a vertical line is a line, it simply has no slope.

d) Parallel lines have equal slopes, so solve $f'(x) = 15$ : $3x^2 - 6x - 9 = 15$ , $3x^2 - 6x - 24 = 0$ , $x^2 - 2x - 8 = 0$ , $(x - 4)(x + 2) = 0$ . So $x = 4$ or $x = -2$ , with $f(4) = 64 - 48 - 36 + 5 = -15$ and $f(-2) = -8 - 12 + 18 + 5 = 3$ . The points are $(4, -15)$ and $(-2, 3)$ . The question asks for POINTS, so both coordinates are expected; stopping at $x = 4$ and $x = -2$ loses part of the marks.

e) Complete the square: $3x^2 - 6x - 9 = 3(x^2 - 2x + 1) - 3 - 9 = 3(x - 1)^2 - 12$ . Since $(x - 1)^2 \ge 0$ , every slope satisfies $f'(x) \ge -12$ , with equality only at $x = 1$ . The equation $f'(x) = -13$ would need $3(x - 1)^2 = -1$ , impossible: no tangent has slope $-13$ . The smallest slope of a tangent is $-12$ , reached at $x = 1$ , which is why the tangent of a) looked like the steepest descent on the figure. Every question about the slopes of a curve is a question about the FUNCTION $f'$ , here a parabola that opens upward.

Exercise 5: Tangent lines through a point that is not on the curve

When the point $P$ is ON the curve, its abscissa is the point of tangency and the tangent line is immediate. When $P$ is NOT on the curve, the point of tangency is unknown: call its abscissa $a$ , write the tangent at $a$ with $a$ as a letter, and require that this line passes through $P$ . That condition is an equation in $a$ , and each of its solutions gives one tangent.

The figure shows the parabola $y = x^2$ and the point $P(1, -3)$ , which lies below it.

a) Show that the tangent to $y = x^2$ at the point of abscissa $a$ has equation $y = 2ax - a^2$ .
b) Find the equations of all the tangents to $y = x^2$ that pass through $P(1, -3)$ , and their points of tangency.
c) Show that no tangent to $y = x^2$ passes through $Q(0, 1)$ . More generally, how many tangents pass through a point $(p, q)$ , depending on $q$ and $p^2$ ?
d) Find all the tangents to $y = x^3$ that pass through $R(2, 0)$ .
e) A student answers b) with the line through $P$ of slope $f'(1) = 2$ , that is $y = 2x - 5$ . Show that this line does not even touch the parabola.

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Answers

a) $y = a^2 + 2a(x - a) = 2ax - a^2$
b) $y = 6x - 9$ , tangent at $(3, 9)$ ; $y = -2x - 1$ , tangent at $(-1, 1)$
c) $a^2 = -1$ has no solution; two tangents if $q < p^2$ , one if $q = p^2$ , none if $q > p^2$
d) $y = 0$ (at the origin) and $y = 27x - 54$ (at $(3, 27)$ )
e) $x^2 = 2x - 5$ gives $x^2 - 2x + 5 = 0$ , discriminant $-16 < 0$ : no common point

a) With $f(x) = x^2$ , $f'(x) = 2x$ , so at the point $(a, a^2)$ the slope is $2a$ and the tangent is $y = a^2 + 2a(x - a) = 2ax - 2a^2 + a^2 = 2ax - a^2$ . Here $a$ is kept as a LETTER on purpose: this equation describes every tangent to the parabola at once, and the next questions choose $a$ .

b) The tangent at $a$ passes through $P(1, -3)$ when $-3 = 2a \cdot 1 - a^2$ , that is $a^2 - 2a - 3 = 0$ , $(a - 3)(a + 1) = 0$ . For $a = 3$ : $y = 6x - 9$ , tangent at $(3, 9)$ . For $a = -1$ : $y = -2x - 1$ , tangent at $(-1, 1)$ . Check that both lines pass through $P$ : $6 - 9 = -3$ and $-2 - 1 = -3$ . The figure of the solution shows the two lines leaving $P$ and grazing the parabola on either side. Two tangents is the normal situation for a point below a parabola that opens upward.

c) Through $Q(0, 1)$ the condition is $1 = 2a \cdot 0 - a^2$ , that is $a^2 = -1$ : no real solution, so no tangent passes through $Q$ . In general the tangent at $a$ passes through $(p, q)$ when $q = 2ap - a^2$ , that is $a^2 - 2pa + q = 0$ , a quadratic in $a$ with discriminant $4p^2 - 4q = 4(p^2 - q)$ . If $q < p^2$ , the point is below the parabola and there are two tangents; if $q = p^2$ , it is on the parabola and there is one; if $q > p^2$ , it is inside the cup and there are none. $P$ has $-3 < 1$ , $Q$ has $1 > 0$ : the discriminant sorts the points, and the figure makes it believable, since from inside the cup every line meets the parabola twice.

d) For $y = x^3$ , $y' = 3x^2$ , and the tangent at $a$ is $y = a^3 + 3a^2(x - a) = 3a^2x - 2a^3$ . Through $R(2, 0)$ : $0 = 6a^2 - 2a^3 = 2a^2(3 - a)$ , so $a = 0$ or $a = 3$ . For $a = 0$ the tangent is $y = 0$ , the $x$ -axis, tangent at the origin; for $a = 3$ it is $y = 27x - 54$ , tangent at $(3, 27)$ , and $27 \cdot 2 - 54 = 0$ confirms that it passes through $R$ . The tangent $y = 0$ CROSSES the curve at the origin, since $x^3$ changes sign there: a tangent line is defined by its slope, not by staying on one side of the curve.

e) The student used the slope at the abscissa of $P$ , as if $P$ were the point of tangency. But $P$ is not on the parabola, since $1^2 = 1 \ne -3$ , so $f'(1)$ is the slope of the curve at $(1, 1)$ , a point that has nothing to do with the line through $P$ . The line $y = 2x - 5$ meets the parabola where $x^2 = 2x - 5$ , that is $x^2 - 2x + 5 = 0$ , whose discriminant is $4 - 20 = -16 < 0$ : no common point at all. A tangent that does not touch the curve is the clearest sign that the point of tangency was never looked for.

Exercise 6: Higher derivatives: the product rule applied twice

The second derivative $f''$ is the derivative of $f'$ , the third is the derivative of $f''$ , and $f^{(n)}$ denotes the $n$ -th. Nothing new is needed to compute them, only care: each step differentiates the RESULT of the previous one, simplified first. Two facts organize the chapter: a polynomial of degree $n$ has $f^{(n)}$ constant and $f^{(n+1)} = 0$ , and the second derivative of a product is not the product of the second derivatives.

When a pattern appears after three or four derivatives, the general formula is a conjecture until it is proved by induction.

a) Let $f(x) = x^4 - 2x^3 + x$ . Find $f'$ , $f''$ , $f'''$ , $f^{(4)}$ and $f^{(5)}$ , and compute $f''(2)$ .
b) Apply the product rule twice to prove that $(uv)'' = u''v + 2u'v' + uv''$ . Use it to find the second derivative of $x^2e^x$ , and explain why $(uv)'' = u''v''$ is false.
c) Let $g(x) = xe^x$ . Compute $g'$ , $g''$ and $g'''$ , conjecture $g^{(n)}$ , and prove your conjecture by induction. What is $g^{(n)}(0)$ ?
d) Let $h(x) = \frac{1}{x}$ . Find $h'$ , $h''$ , $h'''$ , conjecture and prove a formula for $h^{(n)}(x)$ , and compute $h^{(5)}(1)$ .
e) Find the $100$ -th derivative of $P(x) = 5x^{100} + x^{99} - 7x^3$ , and the first derivative of $x^7$ that is identically zero.

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Answers

a) $4x^3 - 6x^2 + 1$ , $12x^2 - 12x$ , $24x - 12$ , $24$ , $0$ ; $f''(2) = 24$
b) $(x^2e^x)'' = (x^2 + 4x + 2)e^x$ ; at $x = 1$ , $u''v'' = 2e \ne 7e$
c) $g^{(n)}(x) = (x + n)e^x$ , so $g^{(n)}(0) = n$
d) $h^{(n)}(x) = \frac{(-1)^n n!}{x^{n+1}}$ , $h^{(5)}(1) = -120$
e) $P^{(100)}(x) = 5 \cdot 100!$ ; the $8$ -th derivative of $x^7$

a) $f'(x) = 4x^3 - 6x^2 + 1$ , $f''(x) = 12x^2 - 12x$ , $f'''(x) = 24x - 12$ , $f^{(4)}(x) = 24$ and $f^{(5)}(x) = 0$ . Each derivative lowers the degree by one, from $4$ to $0$ , then the constant $24$ differentiates to $0$ and every later derivative stays $0$ . At $x = 2$ : $f''(2) = 48 - 24 = 24$ . Note that the constant term of $f$ and the term $x$ both vanish by the second derivative: information is lost at each step, which is why a higher derivative never determines the function.

b) First derivative by the product rule: $(uv)' = u'v + uv'$ . Differentiate again, applying the product rule to EACH of the two products: $(u'v)' = u''v + u'v'$ and $(uv')' = u'v' + uv''$ . Adding, $(uv)'' = u''v + 2u'v' + uv''$ . For $u = x^2$ , $v = e^x$ : $u' = 2x$ , $u'' = 2$ , $v = v' = v'' = e^x$ , so $(x^2e^x)'' = 2e^x + 2 \cdot 2x \cdot e^x + x^2e^x = (x^2 + 4x + 2)e^x$ . Check against the first derivative $x(x + 2)e^x = (x^2 + 2x)e^x$ of Exercise 2: its derivative is $(2x + 2)e^x + (x^2 + 2x)e^x = (x^2 + 4x + 2)e^x$ . The formula $u''v''$ gives $2e^x$ : at $x = 0$ it happens to agree with the true value $2$ , but at $x = 1$ it gives $2e$ against $7e$ . Missing the middle term $2u'v'$ is the error, and the correct formula has the same coefficients $1, 2, 1$ as $(a + b)^2$ .

c) Product rule: $g'(x) = e^x + xe^x = (x + 1)e^x$ . Again: $g''(x) = e^x + (x + 1)e^x = (x + 2)e^x$ , and $g'''(x) = (x + 3)e^x$ . Conjecture: $g^{(n)}(x) = (x + n)e^x$ for $n \ge 1$ . Induction. Base: $g'(x) = (x + 1)e^x$ . Step: if $g^{(n)}(x) = (x + n)e^x$ , then by the product rule $g^{(n+1)}(x) = 1 \cdot e^x + (x + n)e^x = (x + n + 1)e^x$ , which is the formula for $n + 1$ . Hence $g^{(n)}(x) = (x + n)e^x$ for every $n \ge 1$ , and $g^{(n)}(0) = n$ . Stopping after three derivatives and writing the pattern is a guess, and a marker gives the proof marks only for the inductive step written out.

d) Rewrite $h(x) = x^{-1}$ . Then $h'(x) = -x^{-2}$ , $h''(x) = 2x^{-3}$ , $h'''(x) = -6x^{-4}$ . The signs alternate, the coefficients are $1, 2, 6 = 3!$ , the exponent is $-(n + 1)$ : conjecture $h^{(n)}(x) = (-1)^n n!\,x^{-(n+1)}$ . Induction: true for $n = 1$ , since $(-1)^1 \cdot 1! \cdot x^{-2} = -x^{-2}$ ; if true for $n$ , the power rule gives $h^{(n+1)}(x) = (-1)^n n! \cdot (-(n + 1))x^{-(n+2)} = (-1)^{n+1}(n + 1)!\,x^{-(n+2)}$ , the formula for $n + 1$ . So $h^{(n)}(x) = \frac{(-1)^n n!}{x^{n+1}}$ , and $h^{(5)}(1) = (-1)^5 \cdot 120 = -120$ . Without rewriting $\frac{1}{x}$ as a power, each derivative would need the quotient rule and the pattern would be buried under the algebra.

e) The term $x^{99}$ and the term $-7x^3$ are polynomials of degree less than $100$ , so their $100$ -th derivative is $0$ . For $x^{100}$ , each derivative brings down the current exponent: $100 \cdot 99 \cdot 98 \cdots 1 = 100!$ , a constant. Hence $P^{(100)}(x) = 5 \cdot 100!$ . For $x^7$ , the $7$ -th derivative is the constant $7! = 5040$ and the $8$ -th is the first one that is identically zero. Answering $0$ for $P^{(100)}$ because $100$ derivatives feels like many is the error: the degree decides, and here the degree is exactly $100$ .

Exercise 7: Product and quotient rules from a table of values and from a graph

The product and quotient rules need only four numbers at the point: the values of the two functions and of their derivatives. A function can therefore be known only through a table, or through its graph, and the derivative of a product or a quotient still be computed exactly. The table gives two functions $f$ and $g$ , differentiable everywhere:

$\begin{array}{c|cccc} x & f(x) & f'(x) & g(x) & g'(x) \\ \hline 1 & 2 & 5 & 1 & -3 \\ 2 & 3 & -1 & -2 & 4 \end{array}$

The figure gives the graphs of two other functions $u$ (blue) and $v$ (orange), each made of line segments joining the marked points.

a) Let $P = fg$ and $Q = \frac{f}{g}$ . Find $P'(2)$ and $Q'(2)$ .
b) Find the derivatives at $x = 1$ of $R(x) = x^2f(x)$ , $S(x) = e^xg(x)$ and $T(x) = f(x)^2$ , the last one as the product $f \cdot f$ .
c) Find the equation of the tangent line to the graph of $P$ at $x = 2$ .
d) Let $U = uv$ and $W = \frac{u}{v}$ . Read what you need on the figure and find $U'(1)$ , $U'(4)$ and $W'(4)$ .
e) Can the product rule give $U'(2)$ ? Compute the slopes of $U$ just to the left and just to the right of $x = 2$ , and conclude.

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Answers

a) $P'(2) = 14$ , $Q'(2) = -\frac{5}{2}$
b) $R'(1) = 9$ , $S'(1) = -2e$ , $T'(1) = 20$
c) $y = 14x - 34$
d) $U'(1) = 6$ , $U'(4) = -4$ , $W'(4) = -\frac{4}{25}$
e) No, $u'(2)$ does not exist; left slope $6$ , right slope $-3$ : $U'(2)$ does not exist

a) Product rule at $x = 2$ : $P'(2) = f'(2)g(2) + f(2)g'(2) = (-1)(-2) + 3 \cdot 4 = 2 + 12 = 14$ . Quotient rule: $Q'(2) = \frac{f'(2)g(2) - f(2)g'(2)}{g(2)^2} = \frac{(-1)(-2) - 3 \cdot 4}{(-2)^2} = \frac{2 - 12}{4} = -\frac{5}{2}$ . Two slips cost these marks: reading a value from the wrong row of the table, and squaring $g(2) = -2$ into $-4$ instead of $4$ . Write the rule with the letters first, then substitute, one number per letter.

b) $R'(x) = 2xf(x) + x^2f'(x)$ , so $R'(1) = 2 \cdot 2 + 1 \cdot 5 = 9$ . $S'(x) = e^xg(x) + e^xg'(x) = e^x(g(x) + g'(x))$ , so $S'(1) = e(1 - 3) = -2e$ , an exact answer that stays in terms of $e$ . $T = f \cdot f$ gives $T'(x) = f'(x)f(x) + f(x)f'(x) = 2f(x)f'(x)$ , so $T'(1) = 2 \cdot 2 \cdot 5 = 20$ . The tempting $T'(1) = f'(1)^2 = 25$ is the factor-by-factor rule again; the square of a function is a product, and the product rule is the only tool of this chapter that handles it.

c) The tangent to the graph of $P$ at $x = 2$ goes through $(2, P(2))$ with slope $P'(2)$ . From the table, $P(2) = f(2)g(2) = 3 \cdot (-2) = -6$ , and $P'(2) = 14$ from a). So $y = -6 + 14(x - 2) = 14x - 34$ . The value $P(2)$ is not in the table and must be computed from it; using $f(2) = 3$ as the height of the point is a frequent slip that shifts the whole line.

d) Near $x = 1$ , $u$ is the segment from $(0, 1)$ to $(2, 5)$ : $u(1) = 3$ and $u'(1) = \frac{5 - 1}{2 - 0} = 2$ ; $v$ is horizontal there, $v(1) = 3$ and $v'(1) = 0$ . So $U'(1) = u'(1)v(1) + u(1)v'(1) = 2 \cdot 3 + 3 \cdot 0 = 6$ . Near $x = 4$ , $u$ is the segment from $(2, 5)$ to $(7, 0)$ , slope $-1$ , so $u(4) = 3$ and $u'(4) = -1$ ; $v$ is the segment from $(3, 3)$ to $(7, 1)$ , slope $-\frac{1}{2}$ , so $v(4) = \frac{5}{2}$ and $v'(4) = -\frac{1}{2}$ . Then $U'(4) = (-1) \cdot \frac{5}{2} + 3 \cdot \left(-\frac{1}{2}\right) = -4$ and $W'(4) = \frac{u'(4)v(4) - u(4)v'(4)}{v(4)^2} = \frac{-\frac{5}{2} + \frac{3}{2}}{\frac{25}{4}} = -\frac{4}{25}$ . The slopes are computed between two MARKED points, never from a value read between grid lines.

e) No: the product rule needs $u'(2)$ and $v'(2)$ , and $u$ has a corner at $x = 2$ , where its slope jumps from $2$ to $-1$ , so $u'(2)$ does not exist and the rule cannot be quoted. On $[0, 3]$ , $v = 3$ is constant, so $U = 3u$ near $x = 2$ : its slope is $3 \cdot 2 = 6$ just to the left and $3 \cdot (-1) = -3$ just to the right. The two one-sided slopes differ, so $U$ has a corner at $x = 2$ and $U'(2)$ does not exist. A rule is a statement with hypotheses, both factors differentiable at the point; when one of them fails, look at the function itself.

Exercise 8: Five statements to correct

Each statement below was written by a student on a MATH 140 assignment, and each is false. Say what is wrong, give a counterexample with numbers, and write the correct statement.

a) If $f$ and $g$ are differentiable, then $(fg)' = f'g'$ .
b) By the power rule, $\frac{d}{dx}e^x = xe^{x-1}$ .
c) $\frac{d}{dx}\left(e^3\right) = 3e^2$ and $\frac{d}{dx}\left(\sqrt{\pi}\right) = \frac{1}{2\sqrt{\pi}}$ .
d) The quotient rule is $\left(\frac{f}{g}\right)' = \frac{fg' - f'g}{g^2}$ .
e) At a point where $f'(a) = 0$ , the normal line would have slope $-\frac{1}{0}$ , so the curve has no normal line there.

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Answers

a) False: $(x \cdot x)' = 2x$ , not $1$ . $(fg)' = f'g + fg'$ .
b) False: the slope of $e^x$ at $0$ is $1$ , not $0$ . $(e^x)' = e^x$ .
c) False: $e^3$ and $\sqrt \pi$ are constants, their derivatives are $0$ .
d) False: for $f = x$ , $g = 1$ it gives $-1$ . $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ .
e) False: the normal is the vertical line $x = a$ .

a) FALSE. Take $f(x) = g(x) = x$ : $f'g' = 1 \cdot 1 = 1$ , while $fg = x^2$ has derivative $2x$ , which is $6$ at $x = 3$ , not $1$ . Correct statement: $(fg)' = f'g + fg'$ , which gives $1 \cdot x + x \cdot 1 = 2x$ . The factor-by-factor rule is the single most frequent error of the chapter, and the quickest check against it is a product of two polynomials, where expanding gives the answer independently.

b) FALSE. The power rule is about $x^n$ : a VARIABLE base raised to a CONSTANT exponent. In $e^x$ the base is the constant $e$ and the exponent is the variable: it is an exponential function, not a power. The student's formula gives $0 \cdot e^{-1} = 0$ at $x = 0$ , a horizontal tangent, while the graph of $e^x$ is rising there with slope $1$ . Correct statement: $\frac{d}{dx}e^x = e^x$ , and $e$ is precisely the base for which the slope at $0$ equals $1$ .

c) FALSE. $e^3 \approx 20$ and $\sqrt \pi \approx 1.77$ are NUMBERS: their graphs, as functions of $x$ , are horizontal lines, with slope $0$ . The student differentiated with respect to $e$ and to $\pi$ , which are not variables. Correct statement: the derivative of any constant is $0$ ; only an expression containing $x$ has a nonzero derivative. The same trap hides inside longer expressions: $(x^2 + \pi^2)' = 2x$ , not $2x + 2\pi$ .

d) FALSE, the two terms of the numerator are swapped. With $f(x) = x$ and $g(x) = 1$ , the quotient is $x$ , whose derivative is $1$ ; the student's formula gives $\frac{x \cdot 0 - 1 \cdot 1}{1} = -1$ , the opposite. Correct statement: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ , derivative of the top FIRST. Swapping the order changes the sign of every answer, so a mnemonic is worth having: low d-high minus high d-low, over the square of what is below.

e) FALSE. When $f'(a) = 0$ , the tangent at $a$ is the horizontal line $y = f(a)$ , and the line perpendicular to it through $(a, f(a))$ is the VERTICAL line $x = a$ . It exists; it just has no slope, which is why the formula $-\frac{1}{f'(a)}$ cannot produce it. For $f(x) = x^2$ at $a = 0$ : tangent $y = 0$ , normal $x = 0$ , the $y$ -axis. Correct statement: if $f'(a) \ne 0$ the normal has slope $-\frac{1}{f'(a)}$ ; if $f'(a) = 0$ the normal is $x = a$ .

Exercise 9: A particle on a line: velocity, acceleration, and the distance travelled

A particle moves along a straight line, and its position at time $t$ seconds is $s(t) = t^3 - 9t^2 + 24t$ metres, for $0 \le t \le 6$ . Its velocity is $v(t) = s'(t)$ and its acceleration is $a(t) = v'(t) = s''(t)$ . The SIGN of $v$ gives the direction of motion; the particle speeds up when $v$ and $a$ have the same sign and slows down when they have opposite signs, because the speed is $|v|$ , not $v$ .

The figure shows the graph of the position. The acceleration here is NOT constant, so the constant-acceleration formulas of a physics course do not apply: everything comes from $s'$ and $s''$ .

a) Find $v(t)$ and $a(t)$ . When is the particle at rest?
b) On which intervals does the particle move in the positive direction, and in the negative direction? Find its positions at $t = 0, 2, 4, 6$ .
c) Find the displacement and the total distance travelled over $[0, 6]$ .
d) Draw a sign chart of $v$ and $a$ , and find when the particle speeds up and when it slows down.
e) Write $v(t) = 3(t - 3)^2 - 3$ . A student concludes that the particle is slowest at $t = 3$ . Correct this, and give the average velocity and the average speed over $[0, 6]$ .

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Answers

a) $v(t) = 3t^2 - 18t + 24 = 3(t - 2)(t - 4)$ , $a(t) = 6t - 18$ ; at rest at $t = 2$ s and $t = 4$ s
b) Positive on $[0, 2)$ and $(4, 6]$ , negative on $(2, 4)$ ; $s = 0, 20, 16, 36$ m
c) Displacement $36$ m, distance $20 + 4 + 20 = 44$ m
d) Speeds up on $(2, 3)$ and $(4, 6]$ , slows down on $[0, 2)$ and $(3, 4)$
e) At $t = 3$ the VELOCITY is smallest ( $-3$ m/s), the speed is $3$ m/s; slowest at $t = 2$ and $4$ . Averages $6$ m/s and $\frac{22}{3}$ m/s

a) By the power rule, $v(t) = s'(t) = 3t^2 - 18t + 24$ m/s, and $a(t) = v'(t) = 6t - 18$ m/s $^2$ . Factor the velocity: $v(t) = 3(t^2 - 6t + 8) = 3(t - 2)(t - 4)$ . The particle is at rest when $v(t) = 0$ : at $t = 2$ s and $t = 4$ s. On the figure these are the two instants where the position graph has a horizontal tangent. At rest does not mean without acceleration: $a(2) = -6$ and $a(4) = 6$ , which is exactly why the particle does not stay at rest.

b) The factored form gives the sign of $v$ at once: both factors negative on $[0, 2)$ , so $v > 0$ ; opposite signs on $(2, 4)$ , so $v < 0$ ; both positive on $(4, 6]$ , so $v > 0$ . The particle moves in the positive direction on $[0, 2)$ and $(4, 6]$ , in the negative direction on $(2, 4)$ . Positions: $s(0) = 0$ , $s(2) = 8 - 36 + 48 = 20$ , $s(4) = 64 - 144 + 96 = 16$ , $s(6) = 216 - 324 + 144 = 36$ metres. The particle goes out to $20$ m, comes back to $16$ m, then leaves again to $36$ m.

c) The displacement is $s(6) - s(0) = 36$ m. The distance counts every leg positively, split at the changes of direction found in b): $|s(2) - s(0)| + |s(4) - s(2)| + |s(6) - s(4)| = 20 + 4 + 20 = 44$ m. Answering $36$ m for the distance ignores the backward leg, which is travelled once backward and then once more forward: the $4$ m between $16$ and $20$ are covered three times in all. The split points are the zeros of $v$ , never the endpoints alone.

d) $a(t) = 6(t - 3)$ is negative on $[0, 3)$ and positive on $(3, 6]$ . Combining with the sign of $v$ , interval by interval: on $[0, 2)$ , $v > 0$ and $a < 0$ , opposite signs, the particle slows down; on $(2, 3)$ , $v < 0$ and $a < 0$ , same sign, it speeds up (backward); on $(3, 4)$ , $v < 0$ and $a > 0$ , it slows down; on $(4, 6]$ , $v > 0$ and $a > 0$ , it speeds up. The sign chart of the solution puts the four sign changes in one picture. Deciding from $a$ alone (speeding up because $a > 0$ on $(3, 4)$ ) is the classic error: on $(3, 4)$ the particle moves backward, and a positive acceleration there brakes it.

e) Completing the square, $3t^2 - 18t + 24 = 3(t^2 - 6t + 9) - 27 + 24 = 3(t - 3)^2 - 3$ , so $v(t) \ge -3$ with equality at $t = 3$ : the VELOCITY is smallest at $t = 3$ , where the particle moves backward at its greatest speed, $|v(3)| = 3$ m/s. It is the SPEED $|v|$ that measures how slow the particle is, and the speed is $0$ , its smallest possible value, at $t = 2$ and $t = 4$ . The average velocity is the displacement over the time, $\frac{36}{6} = 6$ m/s; the average speed is the distance over the time, $\frac{44}{6} = \frac{22}{3}$ m/s. They differ because the particle turned back once, and they would agree only for a motion that never changes direction.

Exercise 10: A final exam question: the curve y = e^x / x

Let $f(x) = \frac{e^x}{x}$ , defined for $x \ne 0$ ; its graph is shown below. This is the shape of a long final exam question on the chapter: a quotient rule, a factored derivative, horizontal tangents and normals, a tangent at a given point, a tangent through a point off the curve, and a second derivative, every rule named and no calculator.

Answers are exact: keep $e$ , $e^2$ and $e^{-1}$ as they are.

a) Show that $f'(x) = \frac{e^x(x - 1)}{x^2}$ .
b) Find the points where the tangent is horizontal, and the normal line at each of them.
c) Find the equation of the tangent line at $x = -1$ , and where it crosses the $x$ -axis.
d) Find all the tangent lines to the curve that pass through the origin.
e) Show that $f''(x) = \frac{e^x(x^2 - 2x + 2)}{x^3}$ , and deduce that $f''(x)$ is never $0$ .

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Answers

a) $f'(x) = \frac{xe^x - e^x}{x^2} = \frac{e^x(x - 1)}{x^2}$
b) Only $(1, e)$ ; the normal there is $x = 1$
c) $y = -\frac{2x + 3}{e}$ , crossing at $x = -\frac{3}{2}$
d) One tangent: $y = \frac{e^2}{4}x$ , at $\left(2, \frac{e^2}{2}\right)$
e) $x^2 - 2x + 2 = (x - 1)^2 + 1 > 0$ and $e^x > 0$ : $f''(x) \ne 0$

a) Quotient rule with top $e^x$ , whose derivative is $e^x$ , and bottom $x$ , whose derivative is $1$ : $f'(x) = \frac{e^x \cdot x - e^x \cdot 1}{x^2} = \frac{e^x(x - 1)}{x^2}$ , for $x \ne 0$ . Factoring $e^x$ is part of the answer: it is what makes the sign and the zeros of $f'$ readable in the next questions.

b) A horizontal tangent needs $f'(x) = 0$ . Since $e^x > 0$ and $x^2 > 0$ on the domain, $f'(x) = 0$ exactly when $x - 1 = 0$ : $x = 1$ , and $f(1) = e$ . The only point with a horizontal tangent is $(1, e)$ , the bottom of the right branch on the figure; the left branch has none, since there $x - 1 < 0$ makes $f' < 0$ . At $(1, e)$ the tangent is $y = e$ and the normal is the vertical line $x = 1$ .

c) At $x = -1$ : $f(-1) = \frac{e^{-1}}{-1} = -\frac{1}{e}$ and $f'(-1) = \frac{e^{-1}(-2)}{1} = -\frac{2}{e}$ . The tangent is $y = -\frac{1}{e} - \frac{2}{e}(x + 1) = -\frac{2x + 3}{e}$ . It crosses the $x$ -axis where $2x + 3 = 0$ , at $x = -\frac{3}{2}$ . Keeping $e$ exact makes the line clean; a decimal slope such as $-0.74$ would break the no-calculator rule and hide the fact that the intercept is exactly $-\frac{3}{2}$ .

d) The origin is not on the curve, since $f(0)$ does not exist, so the point of tangency is unknown: call its abscissa $a \ne 0$ . The tangent at $a$ is $y = f(a) + f'(a)(x - a)$ , and it passes through $(0, 0)$ when $0 = f(a) - af'(a) = \frac{e^a}{a} - \frac{e^a(a - 1)}{a} = \frac{e^a(2 - a)}{a}$ . Since $e^a \ne 0$ , $a = 2$ . Then $f(2) = \frac{e^2}{2}$ and $f'(2) = \frac{e^2}{4}$ , and the tangent is $y = \frac{e^2}{2} + \frac{e^2}{4}(x - 2) = \frac{e^2}{4}x$ , which does pass through the origin. There is exactly one such tangent, touching the right branch at $\left(2, \frac{e^2}{2}\right)$ .

e) Differentiate $f'(x) = \frac{e^x(x - 1)}{x^2}$ with the quotient rule. The top is a product, so its derivative comes from the product rule: $\left(e^x(x - 1)\right)' = e^x(x - 1) + e^x = xe^x$ . Then $f''(x) = \frac{xe^x \cdot x^2 - e^x(x - 1) \cdot 2x}{x^4} = \frac{xe^x(x^2 - 2x + 2)}{x^4} = \frac{e^x(x^2 - 2x + 2)}{x^3}$ . Completing the square, $x^2 - 2x + 2 = (x - 1)^2 + 1 \ge 1 > 0$ , and $e^x > 0$ , so the numerator never vanishes and $f''(x) \ne 0$ for every $x$ in the domain. Simplifying $f'$ BEFORE differentiating again, with $e^x$ factored out, is what keeps this second derivative to three lines.

Corrected exercises: power, exponential, product and quotient rules (MATH 140)

Course recap

Part A: the basics (/50)

Exercise 1: The power rule, after rewriting roots and fractions as powers

Exercise 2: Product and quotient rules: never factor by factor

Exercise 3: Choosing the form: rewrite first, or use the quotient rule?

Exercise 4: Tangent and normal lines, and where the tangent is horizontal

Exercise 5: Tangent lines through a point that is not on the curve

Part B: problems and reasoning (/50)

Exercise 6: Higher derivatives: the product rule applied twice

Exercise 7: Product and quotient rules from a table of values and from a graph

Exercise 8: Five statements to correct

Exercise 9: A particle on a line: velocity, acceleration, and the distance travelled

Exercise 10: A final exam question: the curve y = e^x / x

See also

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